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3 years ago

What is a example of nuclear fusion

Physics

1 answer:

andrew11 [14]3 years ago

4 0

Energy of our sun. It composed of a lot of hydrogen and by the two fusion of isotopes. Helium is then released.

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A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of

ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= $\frac{\mu_0}{4\pi} \frac{2i}{r}$

i is current , r is distance from wire

B = 10⁻⁷ x $\frac{2\times49}{r}$

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x $\frac{2\times49}{r}$ x 33 x 1

= 3234 x $\frac{10^{-7}}{r}$

This should balance weight of second wire per unit length

3234 x $\frac{10^{-7}}{r}$ = .075

r = $\frac{3234}{.075}$ x 10⁻⁷

= .0043 m

= .43 cm .

5 0

3 years ago

A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve

Gemiola [76]

Answer:

(a) v = 5.42m/s

(b) vo = 4.64m/s

(d) Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

$v=\sqrt{2gh}$ (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

$v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}$

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

$h_{max}=\frac{v_o^2}{2g}$ (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

$v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}$

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}$

$a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}$

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

$\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m$

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0

3 years ago

A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the sam

nikitadnepr [17]

Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

3 0

2 years ago

Ana kicks a stationary 0.2kg soccer ball. The ball leaves her foot at a speed of 20 m/s.

elena55 [62]

Answer:a. 24 kg m/s

b. 3/5s

Explanation:

a.impulse is the change in momentum so at first the momentum is zero because the ball was at rest and the final momentum is 1.2kg*20m/s=24 kg m/s

so the impulse would be (24-0) kg m/s=24 kg m/s

b. so the impulse equation is impulse is force *delts time

so 24 kg m/s=40N*t

t=24 kg m/s /40N=3/5 s

6 0

2 years ago

How did the space rock die??

EastWind [94]

Got shot with a pump shotgun to the head

6 0

2 years ago